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\lhead{MAT\,137\,Y1Y}
\chead{Problem Set \#\,1}
\rhead{Summer 2014}


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\begin{large}
  \noindent
  Name : Robert Staskiewicz \hfill Student Number: 1000340570\\
  \indent  \hfill Tutorial: 0201
\end{large}

\medskip

\noindent
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\subsection*{Topic: Sets, Domains, and Ranges}

\medskip

\begin{enumerate}
 
       
    % place solution to question 1 below

    \item
    \medskip
    \noindent
    Consider the following claim:
    \begin{quote}
    If A, B and C are sets, prove that $B \setminus (A \cup C) = (B \setminus A) \cap (B \setminus C).$
    \end{quote}
    We can express this in the notation of symbolic logic as:
    $$ \ \forall \ Sets \ A,B,C,\ [B \setminus (A \cup C)] = [(B \setminus A) \cap (B \setminus C)] $$
    {\bf Proof:} \\
    \begin{pindent}
		
	\hspace*{-10mm}	We must Prove: $[(B\setminus(A \cup C)) \subseteq ((B \setminus A) \cap (B \setminus C))] \land [((B \setminus A) \cap (B \setminus C)) \subseteq (B\setminus(A \cup C))]$\\
		
        \begin{assumption}{$x \in (B \setminus (A \cup C))$}
		Then $x\in B \land x \notin (A \cup C)$\\
		Then $x \in B \land x \notin A \land x \notin C$ \just{x is not a member of $(A \cup C)$.}\\
		Then $x\in (B \setminus A)$ \just{$x \in B \land x \notin A$}\\
		Then $x\in (B \setminus C)$ \just{$x \in B \land x \notin C$}\\
		Then $x \in  ((B \setminus A) \cap (B \setminus C)) $ \just{$x\in (B \setminus A) \land x\in (B \setminus C)$ }
		\end{assumption}
		Therefore, $(B\setminus(A \cup C)) \subseteq ((B \setminus A) \cap (B \setminus C)) $\\
		
		\begin{assumption}{$x \in  ((B \setminus A) \cap (B \setminus C))$}
				Then $(x \in B \land x \notin A) \land (x \in B \land x \notin C) $ \just{Definition of set intersection.} \\
				Case1:\\
				$x \in B \land x \notin A  $ \\
				Then $x \in (B \setminus (A \cup C)) $\just {x is in B and not in A}\\
				Case2:\\
				$x \in B \land x \notin C$\\
				Then $x \in (B \setminus (A \cup C)) $\just {x is in B and not in C}
		\end{assumption}
		Therefore $ ((B \setminus A) \cup (B \setminus C)) \subseteq (B\setminus(A \cup C))$\\\\
	\hspace*{-10mm}Therefore, $ \ \forall \ Sets \ A,B,C,\ [B \setminus (A \cup C)] = [(B \setminus A) \cap (B \setminus C)] $
    \end{pindent}
    \medskip
    
    % place solution to question 2 below

       
    \item
    \begin{enumerate}
    \item
    Consider the following claim:
       \begin{quote} 
       "A function $f$ is periodic if and only if there exists a $k > 0$ such that for every $x,f(k + x) = f(x)$".\\
       \end{quote}
       We can express this in the notation of symbolic logic as:
       $$    \text{A function is periodic} \iff \exists k \in \R, k>0, \forall x \in \R, f(k+x) = f(x)  $$ \\
       
       \item
       
       \end{enumerate}
       
           Consider the following claim:
           \begin{quote}
           Prove the following by contradiction: If $||x|-|y||<|x-y|, \text{then}\  xy < 0$.
           \end{quote}
       {\bf Proof:} \\
    We can express this in the notation of symbolic logic as:
    $$\forall x,y \in \R,\ (||x|-|y||<|x-y|) \implies (xy<0) $$ \\
		{\bf Proof:} \\
		    \begin{pindent}
		
          \begin{assumption}{$||x|-|y||<|x-y|$ is true}
				Suppose $xy \geq 0$\\
				Then $(x > 0 \land y >0) \lor (x< 0 \land y < 0)$\\\\
				Case1:\\
				If $(x > 0 \land y >0)$\\
				Then $(||x|-|y||<|x-y|) \implies(|x-y|<|x-y|)$\just{We reach a contradiction.}\\\\
				Case2:\\
				If $(x< 0 \land y < 0)$\\
				Then $(||x|-|y||<|x-y|) \implies (|-x+y|<|x-y|)$\\
				\hspace*{10mm}Subcase1:\\
				\hspace*{10mm}If $x>y$\\
				\small\hspace*{10mm}Then $(|-x+y|<|x-y|) \implies (-(-x+y)<(x-y))$\just{$-x+y$ is negative because $x>y$}  \\
				\hspace*{10mm}Then $(|-x+y|<|x-y|) \implies ((x-y)<(x-y))$\just{We reach a contradiction.}\\
				\hspace*{10mm}Subcase2:\\
				\hspace*{10mm}If $y>x$\\
				\hspace*{10mm}Then  $(|-x+y|<|x-y|) \implies ((-x+y)<(-x+y))$\just{We reach a contradiction.}\\
				Therefore, if $||x|-|y||<|x-y|$ is true, then $xy \geq 0 $ cannot be true.\\
          \end{assumption}
          Therefore, $\forall x,y \in \R,\ (||x|-|y||<|x-y|) \implies (xy<0) $
           
      \end{pindent}
      \medskip

	\item
	    \begin{quote}
	     Solve for $x: x^2 - 9x + 20 < 0$
	    \end{quote}
			
			 \begin{assumption}{$x^2-9x+20$}
			 Then $x^2-9x+20=(x-4)(x-5)$ \just{Factoring.}\\
			 Then $(x-4)(x-5)<0 $\just{Roots at $4 \ \text{and} \ 5$.}\\
			 We test the intervals $(-\infty,4),\ (4, 5) ,\ (5, \infty)$\\
			 let $x=0$, then $(0^2-9*0+20=20)$\just{We exclude $(-\infty, 4)$}\\
			 let $x=4.5$, then $(4.5^2)-9(4.5)+20=-0.25$\just{We include $(4,5)$}\\
			 let $x=6$, then $6^2-9(6)+20=2$\just{We exclude $(5, \infty)$}\\
			 Therefore, $\forall x \in \R, ((x^2 - 9x + 20) < 0) \implies \{x\in \R| 4<x<5\} $
			 \end{assumption}

	\item
	\item
	\item
		\begin{enumerate}
		\item
		\item
		 Consider the following function:\\
		 \begin{quote}
		  Domain $=  (-\infty, 1)\cup (1, 3) \cup (3, \infty) $ and a Range $= (5, \infty) $\\\\
		  A function with this Domain and Range is: $ f(x) = (\frac{1}{x^2}+5)\frac{x^2-4x+3}{(x-1)(x-3)} $\\
		  
		  \end{quote}
		  	{\bf Proof:} \\
		  		    \begin{pindent}
		  The Domain appears to be defined at all points but $1$ and $3$.\\
		  We take a function that has an undefined denominator at $x=1$ and $x=3$,
		  so we have $(x-3)$ and $(x-1)$ for a denominator $\frac{1}{(x-1)(x-3)}$. However, we want these points to be undefined, so we make the numerator $(x-3)(x-1)$.
		\end{pindent}
		      \medskip
		\end{enumerate}
\end{enumerate}

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